Algebra precalculus

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I can"t figure out how to lớn simplify \$sin2 heta cos2 heta\$ khổng lồ be \$frac12 sin4 heta \$. I assume double angle identities are involved, but I can"t get it & would lượt thích some help.

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We know the double angle formula for sine is \$sin(2x) = 2sin(x)cos(x)\$.

For convenience, let \$x = 2 heta\$. Then \$4 heta\$ can be written as\$\$4 heta = 2(2 heta) = 2x.\$\$It then follows that\$\$frac12sin(4 heta) = frac12sin(2x) = frac12 cdot 2sin(x)cos(x) = sin(x)cos(x).\$\$But wait--we said \$x = 2 heta\$ ! Plugging this back in allows us to conclude that\$\$frac12 sin(4 heta) = sin(2 heta)cos(2 heta).\$\$

Note:eginequationsin(2x) = 2sin(x)cos(x) \endequation

Therefore we have \$fracsin(4x)2 = fracsin(2(2x))2= frac2sin(2x)cos(2x)2=sin(2x)cos(2x)\$

Does this clarify?

If you are familiar with the double angle identity:

\$sin(2x) = 2sin(x)cos(x)\$ or written another way

\$1 over 2 sin(2x) = sin(x)cos(x)\$

Let \$x = 2 heta\$ & you"ll get:

\$1 over 2 sin(4 heta) = sin(2 heta)cos(2 heta)\$

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